Solve below somewhat symmetric equations:x, y, z subject tox2+y2−xy=3(x−z)2+(y−z)2−(y−z)(x−z)=4(x−z)2+y2−y(x−z)=1x,y,z∈R+

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Volsa280 · Accepted Answer

Step 1We can take a quadrilateral ABCD, for which:AC∩BD={E},&amp;nbsp;AC=BD=z,&amp;nbsp;ED=x,&amp;nbsp;EC=y,&amp;nbsp;∡AEB=60°,&amp;nbsp;AB=2,&amp;nbsp;BC=1&amp;nbsp;and&amp;nbsp;CD=3Thus, we obtain your conditions:x2+y2−xy=ED2+EC2−2ED⋅ECcos⁡60∘=CD2=3,(x−z)2+(y−z)2−(y−z)(x−z)=(z−x)2+(z−y)2−(z−x)(z−y)==EB2+EA2−2EB⋅EAcos⁡60∘=AB2=4and(x−z)2+y2−y(x−z)=(z−x)2+y2+y(z−x)==EB2+EC2−2EB⋅ECcos⁡120∘=1.Now, let BFCD be a parallelogram.Thus,AC=z=BD=FCand∡ACF=∡DEC=60°,which gives that&amp;nbsp;△AFC&amp;nbsp;is equilateral triangle,&amp;nbsp;BA=2,&amp;nbsp;BF=CD=3&amp;nbsp;adn&amp;nbsp;BC=1Now, let R be a rotation around C by&amp;nbsp;−60∘&amp;nbsp;and&amp;nbsp;R(B)=G&amp;nbsp;thus, since&amp;nbsp;△BGC&amp;nbsp;is an equilateral triangle, we obtain:&amp;nbsp;BG=1,&amp;nbsp;FG=R(AB)=2&amp;nbsp;andFG2=22=(3)2+12=BF2+BC2,which says∡FBG=90°,∡FBC=∡FBC+∡GBC=90°+60°=150°,which givesz2=AC2=FC2=FB2+BC2−2FB⋅BCcos⁡150∘=3+1+3=7,whisch givesz=7and from here we get the answer:(x,y,z)=(57,17,7).

Felicity Carter · Accepted Answer

Step 1We multiply the thrid equation by four, subtract it from two, and then substitute with equation one solved for x.[(x−z)2+(y−z)2−(y−z)(x−z)]−4[(x−z)2+y2−y(x−z)]=0x=34−y2+y2⇒y=14−3313z=±3It should be easy to figure out x from here.

Solve below somewhat symmetric equations: x, y, z subject to x^2+y^2 -

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